Page 153 - Schaum's Outline of Theory and Problems of Applied Physics

138 EQUILIBRIUM [CHAP. 11

Hence
−1.2F 1 m + 270 N·m = 0
F 1 = 225 N

The force exerted by the gate on the upper hinge has the same magnitude.
(b) In order that the gate be in translational equilibrium, F 2x must be equal and opposite to F 1 , and F 2y must be
equal and opposite to w. Hence
F 2x = 225 N F 2y = 300 N
Since F 2x and F 2y are perpendicular, the magnitude of F is

2
2
F = F + F = 375 N
2x
2y
If θ is the angle between F 2 and the vertical, as in Fig. 11-12(b),
F 2x
tan θ = = 0.75 θ = 37 ◦
F 2y
The force the gate exerts on the lower hinge is equal and opposite to F 2 , hence it acts downward at an angle of
37 from the vertical and toward the right.
◦
SOLVED PROBLEM 11.12

A 12-ft ladder that weighs 50 lb rests against a frictionless wall at a point 10 ft above the ground. How
much force does the ladder exert (a) on the ground and (b) on the wall?
(a) The forces that act on the ladder are its weight w acting downward from its center, the horizontal reaction force
F 1 of the wall (there is no vertical force component because the wall is frictionless), and the reaction force F 2
of the ground, which has both vertical and horizontal components. Since F 2y and w are the only vertical forces,

F 2y = w = 50 lb
To find F 2x , we begin by finding the value of the angle between the ladder and the ground. From Fig. 11-13
10 ft
sin θ = = 0.833 θ = 56 ◦
12 ft

O
L = (12 ft) (cos q) F 1
1
= (6 ft) (cos q)
L 3
12 ft

q
L 3 L 2 =
10 ft

F 2
F 2y w

q
F 2x

L 1

Fig. 11-13